CODEWITHSUNDEEP
javascriptphpjquerymysqlajax
FutureCake
FutureCake

Reputation: 2953

ajax call fails when i try to create a new table php

Want to send a ajax request to php to create a new table dynamically. I have the following code:

Javascript: 

var file = $('#'+option_file_name.id).prop('files')[0];
var form_data = new FormData(); 
form_data.append('file', file);
form_data.append('column_name', option_file_name.id.slice(12));
send_ajax_request(form_data);

function send_ajax_request(pdata){
    console.log(pdata);
    $.ajax({
        url: "set_points_data.php",
        data: pdata,
        contentType: false,
        processData: false,
        dataType: "json",
        type: 'post',
        success: function (data) {
            console.log(data);
        },
        error: function(a,b,c){
            alert("something went wrong");
        }
    });
}

php script:

in the data.php file i connect to the database and have all the supporting functions. 

<?php
include 'data.php';

$data = null;

if(isset($_FILES)){
    $data = $_FILES;
    $type = (string)$_POST["column_name"];
    create_table($conn, $type);
    return_ajax("succes");
}else{
    return_ajax("error");
}

function create_table($conn, $name){
    send_sql($conn, "CREATE TABLE `".$name."` (
        options varchar(255));"
    );
}
?>

UPDATE:

it turns out that if i change the following code to this the error disappears. see the new code below: 

send_sql($conn, "CREATE TABLE testTable (
    options varchar(255)
);");

but this is not what i want i need to create tables dynamically.

when ever i run run the ajax request i get the error something went wrong. But when i run the php script without the ajax call it works fine. and when i run the php script from the ajax call but without the create table function it also works fine. So i am not sure what is happening here. Hope this is clear if not let me know.

Upvotes: 0

Views: 53

Answers (1)

Celfurion
Celfurion

Reputation: 58

After doing some testing i figured out that ajax is trying to return some data after doing the create table statement. Resulting in a error. if you change it to the following. 

if(isset($_POST)){
    $data = $_FILES;
    $db_name = $_POST["tblname"];
    return_ajax(create_table($conn, $db_name));
}else{
    return_ajax("error");
}

function create_table($conn, $name){
    send_sql($conn, "CREATE TABLE `".$name."` (
        options varchar(255)
    );");
    return "database created";
}

This wil work. not sure what exactly happens but this fixes the ajax error.

Upvotes: 1

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